Question

the distance of a point (-6,y) is 10 units from origin the value of y is​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{The distance of (-6,y) from origin is 10 units}$

$\underline{\textbf{To find:}}$

$\textsf{The value of 'y'}$

$\underline{\textbf{Solution:}}$

$\textsf{The distance of (-6,y) from origin is 10 units}$

$\implies\mathsf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=10}$

$\implies\mathsf{\sqrt{(-6-0)^2+(y-0)^2}=10}$

$\implies\mathsf{\sqrt{36+y^2}=10}$

$\textsf{Squaring on bothsides we get}$

$\mathsf{36+y^2=100}$

$\mathsf{y^2=100-36}$

$\mathsf{y^2=64}$

$\textsf{Taking square root on bothsides,}$

$\implies\boxed{\bf\,y=\pm\,8}$

$\underline{\textbf{Formula used:}}$

$\boxed{\begin{minipage}{7cm} \\\mathsf{The\;distance\;between\;(x_1,y_)\;and\;(x_2,y_2)\;is}\\\\\;\;\;\;\mathsf{d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}\\ \end{minipage}}$

Answer 2

Answer:

+8 and -8

Step-by-step explanation:

Given:- The distance of the point (-6,y) from origin is 10 units.

To Find:- Value of y.

Solution:- The distance of the point (-6,y) from origin(0,0) is 10 units.

As we know,

Distance = $\sqrt{(x_{2}-x_{1} )^{2} + (y_{2}-y_{1} )^{2} }$

⇒ 10 = $\sqrt{(0-( -6 ))^{2} + (0-y)^{2} }$

⇒ 10 = $\sqrt{(6)^{2}+y^{2} }$

⇒ $(10)^{2}$ = $36 + y^{2}$

⇒ $y^{2} = 100 - 36$

⇒ $y=\sqrt{64}$

⇒ $y = ^{}$ ±8

∴ The value of y is ±8.

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